Let M = (a_{ij}),i, j in {1, 2, 3},be a 3 tim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) First,we construct the matrix $M$ based on the condition that $a_{ij} = 1$ if $i$ divides $(j+1)$,and $0$ otherwise.For $i=1$: $j+1$ is divisible

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(A) First,we construct the matrix $M$ based on the condition that $a_{ij} = 1$ if $i$ divides $(j+1)$,and $0$ otherwise.For $i=1$: $j+1$ is divisible by $1$ for $j=1, 2, 3$. Thus,$a_{11}=1, a_{12}=1, a_{13}=1$.For $i=2$: $j+1$ is divisible by $2$ for $j=1, 3$. Thus,$a_{21}=1, a_{22}=0, a_{23}=1$.For $i=3$: $j+1$ is divisible by $3$ for $j=2$. Thus,$a_{31}=0, a_{32}=1, a_{33}=0$.So,$M = \begin{bmatrix} 1 & 1 & 1 \ 1 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix}$.$(A)$ Calculate the determinant: $|M| = 1(0-1) - 1(0-0) + 1(1-0) = -1 + 1 = 0$. Since $|M| = 0$,$M$ is singular and not invertible. ($A$ is false).$(B)$ We solve $M X = -X$,which is $(M + I)X = 0$. $M+I = \begin{bmatrix} 2 & 1 & 1 \ 1 & 1 & 1 \ 0 & 1 & 1 \end{bmatrix}$. The determinant $|M+I| = 2(1-1) - 1(1-0) + 1(1-0) = 0$. Since the determinant is $0$,there exists a non-zero solution $X$. ($B$ is true).$(C)$ The set $\{X \in \mathbb{R}^3 : MX = 0, X 
eq 0\}$ represents the non-trivial null space of $M$. Since $|M| = 0$,the null space is non-trivial. ($C$ is true).$(D)$ $M - 2I = \begin{bmatrix} 1-2 & 1 & 1 \ 1 & 0-2 & 1 \ 0 & 1 & 0-2 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \ 1 & -2 & 1 \ 0 & 1 & -2 \end{bmatrix}$.$|M-2I| = -1(4-1) - 1(-2-0) + 1(1-0) = -3 + 2 + 1 = 0$. Thus,$(M-2I)$ is not invertible. ($D$ is false).Therefore,the correct statements are $B$ and $C$.

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

Similar Questions

The system of equations $kx + y + z = 1$,$x + ky + z = k$,and $x + y + kz = k^2$ has no solution if $k$ is equal to

If $(\alpha, \beta, \gamma)$ is the solution of the system of simultaneous linear equations given by $3x + 4y - 5z = -6$,$2x + 3y - 4z = -7$,and $4x - 2y + z = 9$,then find the value of $\alpha + 3\beta - 2\gamma$.

The system of equations $a + b - 2c = 0$,$2a - 3b + c = 0$,and $a - 5b + 4c = \alpha$ is consistent for $\alpha$ equal to

If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right]$,then the values of $x$ and $y$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module